Points of inflection: f(x) = sqrt(x^2 + 1)

[gpiglover]

The question as to find all relative extrema. Use the second derivatvie test where applicable.

f(x)= ?x^2+1 (all under square root).

Now Im not sure how exactly to even start this derivative. At first I rewrote this problem to (x^2+1)^1/2.

I then did the derivative of that to make it 1/2(x^2+1)^-1/2 *2x (Chain rule). Was I supposed to do the chain rule or did I just make my life difficult?

 

[2000WS6]

Re: Points of inflection

I believe the chain rule applies here.

Now you set what you have for f'(x) = 0 and solve for all values of \(\displaystyle {x}\)

Once you've done that, you have your critical points

Can you handle it from there?

 

[Dr. Flim-Flam]

Re: Points of inflection

What do points of inflection have to do with relative extrema?